Find all y in [0,2π]

a. $\displaystyle 2(cosx)^2-\sqrt{3}\left(\frac{1}{secx}\right) = 0$

$\displaystyle 2cos^2 - \sqrt{3}\left(\frac{\frac{1}{1}{cosx}}\right) = 0$

$\displaystyle 2cos^2 - \sqrt{3} cos x= 0$

How do I progress from here?

b. $\displaystyle cos^2x + cosx = sin^2x$

$\displaystyle cos^2x+cosx-sin^2x=0$

$\displaystyle 2cos^2x+cos=0$

$\displaystyle cosx(2cosx+1)=0$

$\displaystyle cosx=0$

$\displaystyle 2cosx+1=0$

$\displaystyle cosx=\frac{-1}{2}$

$\displaystyle y = \frac\pi2;\frac{3\pi}{2}; \frac{2\pi}{3}; \frac{4\pi}{3}$